the calculus of polynomialsbeachy/courses/229/03fall/calculus.pdf · preface these notes are...

THE CALCULUS

OF POLYNOMIALS

John A. Beachy

Northern Illinois University

1991Revised 1995, 2003

Contents

1 Polynomials 91.1 Roots of polynomials . . . . . . . . . . . . . . . . . . . . . . . . 111.2 Rational roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.3 Approximating real roots . . . . . . . . . . . . . . . . . . . . . . 271.4 Complex roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351.5 Interpolating polynomials . . . . . . . . . . . . . . . . . . . . . . 431.6 Mathematical Induction . . . . . . . . . . . . . . . . . . . . . . . 50

3

4 CONTENTS

Preface

These notes are intended to be used as a supplement to the material usually taught inthe first semester of calculus. Certain techniques used to obtain numerical approx-imations provide the focal point of the notes. Newton’s method uses tangent linesto find successive approximations to solutions of equations. The idea of using atangent line to approximate a function (locally) can be extended to use polynomialsof higher degree. Polynomial approximations are also useful in finding the areabeneath a curve.

This general theme of using polynomials to approximate functions presupposessome knowledge of polynomials. The first chapter of the notes is designed to reviewsome of the necessary background as well as to provide some new information aboutpolynomials. It also includes sections on mathematical induction and the complexnumbers.

The second chapter discusses derivatives of polynomial functions, without usinglimits. (For those already in the know, the tangent line at a point is defined as thelinear part of the Taylor expansion at the point.) The definition of the average valueof a function uses as motivation the statement that is the Fundamental Theorem ofCalculus (in a traditional development). This notion of an average value is then usedto compute areas under curves and to define the integral of a function.

The third chapter introduces the concept of a limit, in connection with sequencesof approximations. Then integrals are reconsidered, using limits of sequences. Itends with some results on infinite series, including a discussion of Taylor series.

To help set the tone of this course, it may be useful to make some generalcomments about mathematics. The field of mathematics involves the constructionand study of abstract models of physical situations. The construction of a modelrequires the selection of explicitly stated and precisely formulated premises. Theseassumptions are called axioms, and the the study of the model then involves drawingconclusions from these fundamental assumptions, using as high a degree of logicalrigor as possible. The rigor of mathematics is not absolute, but is rather in the processof continual development. For example, Euclid’s axiomatization of geometry and his

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6 CONTENTS

study of this model of our spatial surroundings was accepted as completely rigorousfor over two thousand years, but a modern geometer could point out serious flawsin the logical development of the theory.

The choice of the basic assumptions to be taken as axioms usually involves anoversimplification of the facts. For example, models of the U.S. economy cannothope to take into account every variable and still have a workable model. A math-ematical model should only be viewed as the best statement of the known facts. Inmany cases a model should be viewed as merely the most efficient, incorporatingonly enough assumptions to give the desired degree of accuracy in prediction. Forexample, in an area small in comparison to the total surface of the earth, plane ge-ometry gives a good approximation for questions involving relationships of figures.As soon as the problems involve large distances, spherical geometry must be usedas the model. As another example, Newtonian physics is good enough for manyproblems in mechanics, and it is necessary to introduce the additional assumptionsof quantum mechanics only if answers have to be found at the atomic level.

If any distinction at all is to be made between applied mathematics and theoreticalmathematics, it is perhaps most evident when talking about the process of modeling.The applied mathematician is probably more involved in the construction of models,and must ask questions about the efficiency of the models and about how closelythey approximate the real world. The theoretical mathematician is concerned withdeveloping the model, by investigating the implications of the basic assumptions oraxioms. This is done by proving theorems. Of course, if a theorem is proved that isobviously contrary to nature, then it is clear to everyone that the basic assumptionsdo not coincide with reality. The theoretical mathematician is also concerned withthe internal consistency of the models.

In order to make logical deductions from the basic axioms, the language usedmust be extremely precise. This is done by making use of careful definitions, andsymbols that are lifted out of the context of ordinary language in order to strip awayany possible ambiguity. Much of the precision and clarity of mathematics is madepossible by its use of formulas. The modern reader is usually unaware that this isan achievement only of the past few centuries.1 For example, the signs+ and− appeared in manuscripts for the first time in 1481; parentheses first appearedin 1544; brackets and braces appeared essentially for the first time in 1593 in theworks of Vieta; the sign= appeared in 1557; the modern way of writing powerswas first used in 1637 by Descartes, but in 1801 Gauss still wrotexx instead ofx2.

The motivation of the pure mathematician certainly comes partly from the appli-

1The notes on symbols are fromDifferential and Integral Calculus , by Ostrowski.

CONTENTS 7

cations of the theorems that can developed within a given model. But perhaps morethan this, it comes from the joy of creating a theory of particular simplicity, eleganceand broad scope. It is certainly difficult to describe the beauty of a mathematicaltheory, but if one understands the theory, it is not difficult to appreciate its beauty, iffor no other reason than what it shows of the intellectual creativity of human kind.

In defense of the theoretical mathematician, it must be said that a theory shouldnot be judged only on its applicability to presently known problems. The history ofmathematics is filled with examples of particular theories that seemed at the time tobe mere intellectual exercises devoid of any relationship to physical problems, butthat later were discovered to have important applications.

One particularly impressive example is provided by non-Euclidean geometry,which arose from the efforts (extending over two thousand years) to prove thatEuclid’s parallel axiom could be deduced from his other, more obvious, axioms.This seemed to be a matter of interest only to mathematicians. Even Lobacevskii,the founder of the new geometry, was careful to label it “imaginary”, since he couldnot see any meaning for it in the actual world. In spite of this, his ideas laid thefoundation for a new development of geometry, namely the creation of theories ofvarious non-Euclidean spaces. These ideas later became, in the hands of Einstein,the basis of the general theory of relativity, in which the mathematical model consistsof a form of non-Euclidean geometry of four-dimensional space.

The generalizations and abstractions of mathematics often seem at first to bestrange and difficult. But with the very general expansion of knowledge and tech-nology that we are currently experiencing, it becomes necessary to identify andelucidate general underlying principles, in order to tie this information together.The language and concepts of mathematics help to fill this need.

Acknowledgments

These notes were written while teaching the honors calculus sequence duringthe fall of 1986 and the fall of 1991. Some of the material in the first chapter hadbeen covered in previous semesters. In particular, I am indebted to Bill Blair, HenryLeonard, Don McAlister, Linda Sons and Bob Wheeler for their notes on varioustopics.

J.A.B.

8 CONTENTS

Chapter 1

Polynomials

The branch of mathematics that provides methods for the quantitative investigationof various processes of change, motion, and dependence of one quantity on anotheris calledmathematical analysis, or simply analysis. A first course in calculusestablishes some of the basic methods of analysis, done in relatively simple cases.

The development of the methods of analysis was stimulated by problems inphysics. During the 16th century the central problem of physics was the investigationof motion. The expansion of trade, and the accompanying explorations, madeit necessary to improve the techniques of navigation, and these in turn dependedto a large extent on developments in astronomy. In 1543 Copernicus published“On the revolution of the heavenly bodies”, and then the “New astronomy” ofKepler, containing his first and second laws for the motion of planets around thesun, appeared in 1609. The third law was published by Kepler in 1618 in his book“Harmony of the world”. Galileo, on the basis of his study of Archimedes andhis own experiments, laid the foundations for the new mechanics, an indispensablescience for the newly arising technology.

During the Renaissance the Europeans finally became acquainted with Greekmathematics by way of the Arabic translations, after a period of almost one thousandyears of scientific stagnation. The books of Euclid, Ptolemy, and Al-Kharizmi weretranslated in the 12th century from Arabic into Latin, the common scientific languageof Western Europe, and at the same time the earlier Greek and Roman system ofcalculation was gradually replaced by the vastly superior Indian method, which alsoreached the Europeans via the Arabs.

It was not until the 16th century that European mathematicians finally surpassedthe achievements of their predecessors, with the solution by the Italians Tartagliaand Ferrari of the general cubic and fourth degree equations.

The concepts of variable magnitude and function arose gradually as a result

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10 CHAPTER 1. POLYNOMIALS

of the interest of physics in laws of motion, as for example in the work of Keplerand Galileo. Galileo discovered the law of falling bodies by establishing that thedistance fallen increases in direct proportion to the square of the time.

The appearance in 1637 of the new “geometry” of Descartes marked the firstdefinite step toward a mathematics of variable magnitudes. This combined algebraicand geometric techniques, and is now known as analytic geometry. The main contentof the new geometry was the theory of conic sections: the ellipse, hyperbola, andparabola. This theory had been developed extensively by the ancient Greeks ingeometric form, and the combination of this knowledge with algebraic techniquesand the general idea of a variable magnitude produced analytic geometry.

For the Greeks the conic sections were a subject of purely mathematical inter-est, but by the time of Descartes they were of practical importance for astronomy,mechanics, and technology. Kepler discovered that each planet travels around thesun in an elliptical orbit, and Galileo established that an object thrown in the airtravels along a parabolic path (both of these are only first approximations). Thesediscoveries made it necessary to calculate various magnitudes associated with theconic sections and it was the method of Descartes that solved this problem.

The next decisive step was take by Newton and Leibnitz during the second halfof the 16th century, and resulted in the founding of differential and integral calculus.The Greeks and later mathematicians had studied the geometric problems of draw-ing tangents to curves and finding areas and volumes of figures. The remarkablediscovery of the relation of these problems to the problems of the new mechanicsand the formulation of general methods for solving them was brought to completionin the work of Newton and Leibnitz. This relation was discovered because of thepossibility, through the use of analytic geometry, of making a graphical represen-tation of how one variable depends on another. In short, what is involved is theconstruction of a geometric model of relationships involving variable magnitudes.

The simplest relationships are those given by polynomials such asx3− 2x+ 3.

The most elementary ones are the linear polynomials, which have the general formmx+ b, for constantsm andb. Complicated expressions likeex2

− sin3(x) aremuch more difficult to work with than polynomials, and so many times it is usefulto approximate such complicated expressions by using polynomials. The simplestcase would be to attempt to approximate a function by a linear function of the formf (x) = mx+b. At best this is only possible for a small interval ofx values, and sodifferential calculus focuses on the construction and use of tangent lines at variousvalues ofx. By using higher derivatives, the idea of a tangent line can be extendedto the idea of polynomials of higher degree which are “tangent” in some sense to agiven curve. These ideas are introduced in Chapter 2, and provide the motivationfor much of this chapter, which focuses on the algebra of polynomials.

1.1. ROOTS OF POLYNOMIALS 11

1.1 Roots of polynomials

We begin with some notation. The set{0,±1,±2,±3, . . .} is called the set ofintegers. We will use Z for the set of integers, andQ for the set ofrationalnumbers. That is,

Q ={m

n| m,n are integers andn 6= 0

}where we must agree thatm/n and p/q represent the same ratio ifmq = np. Byidentifying the integerm with the fractionm/1, we can think of the setZ as a subsetof Q, and so we can say, very roughly, that we have enlarged the set of integers tothe setQ of rational numbers in order to have a set in which division is possible.

We will use the symbolR for the set ofreal numbers. We will simply viewthem as the set of all decimal numbers. They can be thought of as the coordinatesof the points on a straight line. A precise development of the construction of the setof real numbers requires more sophisticated concepts and much more time than wehave available to us at this point. The rational numbers can be viewed as a subset ofR, since fractions correspond to decimals that are either terminating or repeating.

1.1.1 DEFINITION. An expression of the form

amxm+ am−1xm−1

+ · · · + a1x + a0

is called apolynomial in the indeterminate x. The exponents (and subscripts)m,m-1, . . ., 1, 0 must be non-negative integers, and we will assume that thecoefficientsam, am−1, . . ., a0 are real numbers. We say that an expression of this form is apolynomial over R.

If n is the largest nonnegative integer such thatan 6= 0, then we say that thepolynomial hasdegreen, andan is called theleading coefficientof the polynomial.

According to this definition, the zero polynomial has no degree, and a constantpolynomiala0 has degree 0 whena0 6= 0. It is important to note that two poly-nomials are equal precisely when they have the same degree and all correspondingcoefficients are equal.

If a(x) = amxm+am−1xm−1

+· · ·+a0 andb(x) = bnxn+bn−1xn−1

+· · ·+b0

are polynomials, thena(x) andb(x) can be added by just adding correspondingcoefficients. Their producta(x)b(x) is

a(x)b(x) = ambnxn+m+ · · · + (a0b2+ a1b1+ a2b0)x

2+ (a0b1+ a1b0)x + a0b0.


In the above formula, two polynomials are multiplied by multiplying each termof the first by each term of the second, and then collecting similar terms. Table 1.1.1shows an efficient way to do this, by arranging the similar terms in columns.

Table 1.1.1:(2x4− 5x3

+ 3x + 1)(x2+ 2x − 1)

2x4−5x3

+3x +1× x2

+2x −4

−8x4+20x3

−12x −4+4x5

−10x4+6x2

+2x2x6

−5x5+3x3

+x2

2x6−x5

−18x4+23x3

+7x2−10x −4

We will sometimes need to simplify expressions such as(x + c)n. For n = 2andn = 3 we have the following identities.

(x + c)2 = (x + c)(x + c)

= x(x + c)+ c(x + c)

= x2+ xc+ cx+ c2

= x2+ 2cx+ c2

(x + c)3 = (x + c)(x + c)2

= x(x2+ 2cx+ c2)+ c(x2

+ 2cx+ c2)

= x3+ 2cx2

+ c2x + cx2+ 2c2x + c3

= x3+ 3cx2

+ 3c2x + c3

The coefficients in the above identities can be found from Pascal’s triangle,which is given in Table 1.1.2. In this triangle each row begins and ends with 1, andthe other terms are found by adding together the two numbers immediately abovethe term.

The last row gives the coefficients for(x + c)6. In Section 1.6 we will give aproof of the Binomial Theorem, which computes the coefficients by using a differentformula, and then we will show that we get the same answers either way.

Leta(x) = amxm+am−1xm−1

+· · ·+a0 andb(x) = bnxn+bn−1xn−1

+· · ·+b0

be polynomials. To write down the general formula for the producta(x)b(x), it is


Table 1.1.2: Pascal’s triangle (ton = 6 )

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

1 6 15 20 15 6 1

useful to introduce a notation for sums. It is traditional to use a Greek letter, sigma,to denote a sum. To show the terms that are being added we typically use subscripts,and then part of the notation tells what the subscripts should be. For example, wecan write

a0b2+ a1b1+ a2b0 =

2∑i=0

ai b2−i

to describe the coefficient ofx2 ina(x)b(x). In general, the coefficientck ofa(x)b(x)is given by the formula

ck =

k∑i=0

ai bk−i ,

which can also be written as

ck =∑

i+ j=k

ai b j .

1.1.2 PROPOSITION. If a(x) andb(x) are nonzero polynomials overR, then theproducta(x)b(x) is nonzero and

deg(a(x)b(x)) = deg(a(x))+ deg(b(x)).

Proof. Suppose thata(x) = amxm+ · · · + a0 andb(x) = bnxn

+ · · · + b0, with thedegree ofa(x) = m and the degree ofb(x) = n, so thatam 6= 0 andbn 6= 0. Fromthe general formula for multiplication of polynomials, the leading coefficient ofa(x)b(x) isambn, which must be nonzero since the product of nonzero real numbersis nonzero. Thus the degree ofa(x)b(x) is m+ n sinceambn is the coefficient ofxm+n.


We are interested in solving polynomial equations, or equivalently, in findingroots of polynomials. The main result that we need is thatc is a root of a givenpolynomial p(x) if and only if x − c is a factor ofp(x). To check this we need tobe able to dividep(x) by x − c. The next example should help you recall how todivide polynomials.

Example 1.1.1 ((2x3− 3x2

+ 5x + 1) ÷ (x + 1) ).

To divide the polynomial 2x3− 3x2

+ 5x+ 1 by x+ 1 we can use thestandard algorithm for division, as illustrated in Table 1.1.3.

Table 1.1.3:(2x3− 3x2

+ 5x + 1)÷ (x + 1)

2x2−5x +10

x + 1 2x3−3x2

+5x +12x3

+2x2

−5x2+5x

−5x2−5x

10x +110x +10

−9

Thus 2x3− 3x2

+ 5x + 1= (x + 1)(2x2− 5x + 10)+ (−9). 2

Theorem 1.1.4 will show that for any polynomialp(x), the remainder whenp(x) is divided byx − c is p(c). That is, p(x) = (x − c)q(x) + p(c). Theremainder p(c) andquotient q(x) are unique. The usual proof of this theoremuses the algorithm that we followed in Example 1.1.1. The next lemma leads to asimpler proof.

1.1.3 LEMMA. For any real numberc and any positive integerk, the linear termx − c is a factor ofxk

− ck.


Proof. The proof consists of checking that the following factorization works.

xk− ck= (x − c)(xk−1

+ cxk−2+ · · · + ck−2x + ck−1).

Working with the right hand side, we have

(x − c)(xk−1+ cxk−2

+ · · · + ck−2x + ck−1)

= x(xk−1+cxk−2

+· · ·+ck−2x+ck−1)−c(xk−1+cxk−2

+· · ·+ck−2x+ck−1)

= xk+ cxk−1

− cxk−1+ · · · + ck−1x − ck−1x − ck

= xk− ck ,

and so the factorization is correct.

1.1.4 THEOREM (The remainder theorem). Let p(x) be a nonzero polynomialoverR, and letc be any real number. Then there exists a polynomialq(x) with realcoefficients such that

p(x) = (x − c)q(x)+ p(c).

Moreover, if p(x) = (x − c)q1(x) + k, whereq1(x) is a polynomial overR andkis a real number, thenq1(x) = q(x) andk = p(c).

Proof. If p(x) = amxm+· · ·+a0, thenp(x)−p(c) = am(xm

−cm)+· · ·+a1(x−c).By Lemma 1.1.3,x−c is a factor of each term on the right hand side of the equation,and so it must be a factor ofp(x) − p(c). Thus p(x) − p(c) = (x − c)q(x) forsome polynomialq(x) overR, or equivalently,p(x) = (x − c)q(x)+ p(c).

If p(x) = (x − c)q1(x) + k, then (x − c)(q(x) − q1(x)) = k − p(c). Ifq(x) − q1(x) 6= 0, then by Proposition 1.1.2 the left hand side of the equation hasdegree≥ 1, which contradicts the fact that the right hand side of the equation is aconstant. Thusq(x)− q1(x) = 0, which also implies thatk− p(c) = 0, and so thequotient and remainder are unique.

Given a polynomialp(x) and a real numberc, to find p(c)we simply substitutec in place ofx. We can also refer to this process by saying that weevaluatep(x) atc. When evaluating a polynomial, it is often best to write it with nested parentheses,as in the following example.

In the polynomial2x3− 7x2

+ 5x + 20

we first factorx out of 2x3− 7x2

+ 5x to get

2x3− 7x2

+ 5x + 20= (2x2− 7x + 5)x + 20 .


We then factorx out of 2x2− 7x, giving

2x3− 7x2

+ 5x + 20= ((2x − 7)x + 5)x + 20 .

When entering a polynomial into a graphing calculator, using the nested paren-theses form will sometimes speed up the graphing process, since no exponentialsare used, but just addition, subtraction, and multiplication.

Here is another example, in which we have inserted zeros for the appropriatecoefficients.

x5+ 2x2

− 3 = x5+ 0x4

+ 0x3+ 2x2

+ 0x − 3

= ((((x + 0)x + 0)x + 2)x + 0)x − 3

1.1.5 DEFINITION. Let p(x) = amxm+ · · · + a0 be a polynomial overR. A real

numberc is called aroot of the polynomialp(x) if p(c) = 0.

1.1.6 COROLLARY. Let p(x) be a nonzero polynomial overR, and letc be anyreal number. Thenc if a root of p(x) if and only if x − c is a factor ofp(x).

Proof. Using the Remainder Theorem, we can writep(x) = (x − c)q(x) + p(c),and then it follows thatp(c) = 0 if and only if p(x) = (x − c)q(x).

1.1.7 COROLLARY. A polynomial of degreen has at mostn distinct roots.

Proof. Suppose thatp(x) is a polynomial of degreen. If c is a root ofp(x), thenby Corollary 1.1.6 we can writep(x) = (x− c)q(x), for some polynomialq(x). Ifa is any root ofp(x), then substituting shows that(a− c)q(a) = 0, which impliesthat eitherq(a) = 0 or a = c. This shows that we can reduce the problem to apolynomial of lower degree, sinceq(x) has degreen− 1. If we already know thatq(x) has at mostn−1 distinct roots, thenp(x) can have at mostn−1 distinct rootswhich are different fromc.

One of the facts from geometry is that two distinct points determine a uniquestraight line. In the language of polynomials, this translates into the statement thatif p(x) = a1x + a0 is a polynomial of degree 1, then knowingp(c1) and p(c2) forany two pointsc1 6= c2 completely determines the coefficientsa1 anda0.


The next proposition will show that a similar result holds for quadratic poly-nomials. If p(x) = a2x2

+ a1x + a0 and we are givenp(c1), p(c2), and p(c3),for three distinct pointsc1, c2, andc3, then the coefficientsa2, a1, anda0 are com-pletely determined. In geometric terms, this implies that three points (not on a line)determine a unique parabola.

1.1.8 PROPOSITION. Let a(x) andb(x) be polynomials of degreen. If a(x) andb(x) agree atn+ 1 distinct points, then they must be equal.

Proof. Let p(x) = a(x) − b(x). If a(ci ) = b(ci ) for 1 ≤ i ≤ n + 1, thenp(x)hasn+ 1 distinct roots. Ifp(x) is nonzero, then deg(p(x)) ≤ n because bothp(x)andb(x) have degreen, and so Corollary 1.1.7 shows that this cannot happen. Theonly possibility that is left is thatp(x) is the zero polynomial, which shows thata(x) = b(x).

Example 1.1.2 ((x − 1)3+ 3(x − 1)2+ 3(x − 1)+ 1= x3).

Leta(x) = x3 andb(x) = (x−1)3+3(x−1)2+3(x−1)+1. To showthata(x) andb(x) are equal we only need to show that they agree at fourdistinct points. We will evaluate both polynomials atx = −1, x = 0,x = 1, andx = 2. We havea(−1) = −1, a(0) = 0, a(1) = 1, anda(2) = 8. On the other hand, we haveb(−1) = −8+12−6+1= −1,b(0) = −1+ 3− 3+ 1= 0, b(1) = 1, andb(2) = 1+ 3+ 3+ 1= 8.Thereforex3

= (x − 1)3+ 3(x − 1)2+ 3(x − 1)+ 1. 2

PROBLEMS: Section 1.1

1. Write out the expansions of(x + c)4 and(x + c)5.

2. Add the next row to Pascal’s triangle, as given in the text, and use it to writeout the expansion of(x + 2)7.

3. Rewritex5− 7x3

+ 5x2− 3x + 4 using nested parentheses.

4. Rewritex4+ 5x3

+ 11x2− 2x − 8 using nested parentheses.

5. Use Proposition 1.1.8 to show that the polynomialsa(x) = x3+ x − 4 and

b(x) = (x + 1)3− 3(x + 1)2+ 4(x + 1)− 6 are equal.


6. Use Proposition 1.1.8 to show that the polynomialsa(x) = x3+1 andb(x) =

32(x−1)(x)(x+1)− (x−2)(x)(x+1)+ 1

2(x−2)(x−1)(x+1) are equal.

7. Findq(x) with x6− 1 = (x3

− 1)q(x), showing thatx3− 1 is a factor of

x6− 1.

8. Findq(x) with x9− 1 = (x3

− 1)q(x), showing thatx3− 1 is a factor of

x9− 1.

9. Suppose thatmandn are positive integers for whichm is a factor ofn. Explainwhy xm

− 1 must be a factor ofxn− 1.

Hint: Write n = mk and apply Lemma 1.1.3.

10. Explain why rational numbers correspond to decimals that are either repeatingor terminating.

Hint: If q = m/n, then when dividingmbyn to putq into decimal form thereare at mostn different remainders. Conversely, ifd is a repeating decimal,then finds, t such that 10sd − 10td is an integer.

1.2. RATIONAL ROOTS 19

1.2 Rational roots

In this section we will work with polynomials that have integer coefficients. Theeasiest roots to find are those which are rational numbers, and the next propositionshows that there are only a finite number of possibilities that must be checked. Usinga computer program or calculator to graph the polynomial function can help a greatdeal in deciding which possible roots actually work.

1.2.1 PROPOSITION (Rational roots). Let p(x) = amxm+ am−1xm−1

+ · · · +

a1x + a0 be a polynomial with integer coefficients. Ifr/s is a rational root ofp(x)such thatr ands have no factors in common, thenr must be a factor ofa0 andsmust be a factor ofam.

Proof. If p( rs) = 0, then

am(rs)

m+ am−1(

rs)

m−1+ · · · + a1(

rs)+ a0 = 0 ,

then multiplying bysm gives the equation

amr m+ am−1r

m−1s+ · · · + a1rsm−1+ a0sm

= 0 .

This leads to the equations

amr m= s(−am−1r

m−1− · · · − a1rsm−2

− a0sm−1)

anda0sm

= r (−amr m−1− am−1r

m−2s− · · · − a1sm−1) .

Thuss is a factor ofamr m andr is a factor ofa0sm. Sincer ands are assumed tohave no factors in common, we see thatr must be a factor ofa0 ands must be afactor ofam.

Example 1.2.1.

Suppose that we need to find all integer roots of the polynomial

p(x) = x3− 5x2

+ 2x − 10.

Using Proposition 1.2.1, all rational roots ofp(x) can be found bytesting only a finite number of values. By looking at the signs we cansee thatp(x) cannot have any negative roots, so we only need to check


the positive factors of 10. Substituting these factors intop(x)we obtainp(1) = −12, p(2) = −18 andp(5) = 0. Thus 5 is a root ofp(x), andso we can use polynomial division to show thatx3

− 5x2+ 2x− 10=

(x − 5)(x2+ 2). Becausex2

+ 2 has no real roots, this completes thefactorization, and it follows that 5 is the only integer root. 2

Example 1.2.2.

In this example we will find the rational roots of

p(x) = 9x4− 6x3

+ 19x2− 12x + 2.

The possible numerators are the factors of 2, while the possible denom-inators are the factors of 9. The list of possible rational roots is

±1,±2,±1

3,±

2

3,±

1

9,±

2

9.

Graphing the function shows that there is a root between.3 and .4,so this suggests that 1/3 might be a root. (Since the possible rootslie between 0 and 2, on your graphing calculator first use the window0 ≤ x ≤ 2; −1 ≤ y ≤ 1. To take a closer look, you might trythe window 0≤ x ≤ .6; −.2 ≤ y ≤ .2.) If 1/3 is a root, thenwe know thatx − 1/3 must be a factor ofp(x). In order to keepto integer coefficients, it is better to check whether or not 3x − 1 isa factor of p(x). In fact, using polynomial division we can see thatp(x) = (3x− 1)(3x3

− x2+ 6x− 2). Dividing again by 3x− 1 gives

the complete factorization

p(x) = (3x − 1)2(x2+ 2),

and so the only root ofp(x) is 1/3. 2

We need to find some easier methods for evaluating polynomials, and so wewill take a closer look at the method using nested parentheses. The algorithmfor evaluating a polynomial by using nested parentheses is usually calledsyntheticdivision. Suppose thatp(x) = amxm

+am−1xm−1+· · ·+a0 andq(x) = bm−1xm−1

+

bm−2xm−2+ · · · + b0 with p(x) = (x − c)q(x) + p(c). Multiplying this out and

equating coefficients shows that the following equations hold.


am = bm−1

am−1 = bm−2− cbm−1

...

a1 = b0− cb1

a0 = p(c)− cb0

Solving for the coefficients ofq(x) gives

bm−1 = am

bm−2 = bm−1c+ am−1 = amc+ am−1

bm−3 = bm−2c+ am−2 = (amc+ am−1)c+ am−2

...

This shows that the coefficients ofq(x) can be found from the partial answers in the“nested parentheses” procedure for evaluatingp(c).

Example 1.2.3.

We will use “nested parentheses” to dividep(x) = 2x3−7x2

+5x+20by x − 2. As before, we can write

2x3− 7x2

+ 5x + 20= ((2x − 7)x + 5)x + 20 ,

and then we have the following steps:

b2 = 2

b1 = 2 · 2− 7= −3

b0 = (−3) · 2+ 5= −1

p(2) = (−1) · 2+ 20= 18 .

Thus we have

2x3− 7x2

+ 5x + 20= (x − 2)(2x2− 3x − 1)+ 18 . 2


1.2.2 PROPOSITION (Synthetic division).Let p(x) = amxm+am−1xm−1

+· · ·+

a0 be a polynomial with real coefficients. Ifp(x) = (x − c)q(x) + p(c), whereq(x) = bm−1xm−1

+ bm−2xm−2+ · · · + b0, then

bm−1 = am

bm−2 = bm−1c+ am−1

...

bi = bi+1c+ ai+1

...

b0 = b1c+ a1

and p(c) = b0c+ a0.

Example 1.2.4 (The synthetic division algorithm).

In using synthetic division to dividep(x) = amxm+am−1xm−1

+· · ·+a0

by x − c, the work is usually arranged as follows.

c am am−1 am−2 . . .

bm−1c bm−2c . . .

bm−1 bm−2 bm−3 . . .

For example, dividing 2x3−7x2

+5x+20 byx−3 gives the followingtable.

3 2 −7 +5 20+6 −3 6

2 −1 +2 26

We have 2x3− 7x2

− 2x + 20 = (x − 3)(2x2− x + 2) + 26. To

construct the table, bring the 2 down to the bottom row, then multiplyby 3 and insert it as the next term in the second row. As the next step,add it to the corresponding term of of the top row, and then repeat thesesteps, using the answer. 2


In using the synthetic division algorithm, you must insert 0 for any missingpower of x. For example, to dividex5

+ 2x2− 3 by x + 1 we would have the

following steps.

−1 1 0 0 2 0 −3−1 1 −1 −1 1

1 −1 1 1 −1 −2

Thusx5+ 2x2

− 3= (x + 1)(x4− x3+ x2+ x − 1)− 2.

The most compact notation omits the terms in the middle row, as follows.

−1 1 0 0 2 0 −31 −1 1 1 −1 −2

This will be useful later when we need to do some repeated divisions.Using Proposition 1.2.1, all rational roots of the polynomialp(x) can be found

by testing only a finite number of values. We now restrict ourselves even further,and study some ways to find integer roots of equations. The next proposition givesa method due to Newton which can speed up this process considerably.

1.2.3 PROPOSITION. Let p(x) be any polynomial with integer coefficients. Ifcis any integer root ofp(x) andn is any integer, thenc− n must be a factor ofp(n).

Proof. If c is an integer root ofp(x), then it follows from the Remainder Theorem(see Theorem 1.1.4) thatp(x) = (x − c)q(x), which we can rewrite asp(x) =(c− x)(−q(x)). The proof of the Remainder Theorem uses the factorization

xk− ck= (x − c)(xk−1

+ cxk−2+ · · · + ck−2x + ck−1).

If c is an integer, then the coefficients of

xk−1+ cxk−2

+ · · · + ck−2x + ck−1

are certainly integers, and so in the general factorizationp(x) = (x − c)q(x), itfollows that the coefficients ofq(x) are integers.

If we substituten into the equationp(x) = (c − x)(−q(x)), we getp(n) =(c− n)(−q(n)), which shows thatc− n is a factor of the integerp(n).


Example 1.2.5.

The preceding result can be combined with Proposition 1.2.1 to findthe rational roots of equations such asx3

+ 15x2− 3x − 6 = 0. By

Proposition 1.2.1 the possible rational roots are±1,±2,±3,±6. Let-ting p(x) = x3

+ 15x2− 3x − 6 we find thatp(1) = 7. Thus by

Proposition 1.2.3, for any rootc we are guaranteed thatc−1 must be afactor of 7. This eliminates all of the possible values exceptc = 2 andc = −6.

We find thatp(2) = 56, so 2 is not a root. This shows, in addition,thatc− 2 must be a factor of 56 for any rootc, but−6 still passes thistest. Finally,p(−6) = −24, and so this eliminates−6 andp(x) hasno rational roots. 2

If we evaluate a polynomialp(x) at x = c by using synthetic division, wecannot tell whether or notp(c) = 0 until the final step. Our final algorithm usessynthetic division, in reverse. In testing for integer roots, it may be possible toeliminate a potential root after only a few steps of the synthetic division procedure.If the potential integer root passes a divisor test at each step, then it is a root and theanswers at each step provide the coefficients of the quotient.

If p(x) = amxm+am−1xm−1

+· · ·+a0 is a polynomial with integer coefficients,and c is an integer root ofp(x), then p(x) = (x − c)q(x), for a polynomialq(x) = bm−1xm−1

+· · ·+b1x+b0 with integer coefficients. After multiplying thisout, we can solve for the coefficients ofq(x). Solving for−b0 givesb0 = −a0÷ c,We get the following equations, which we might callbackwards synthetic division.

b0 = −a0÷ c

b1 = (b0− a1)÷ c

b2 = (b1− a2)÷ c...

Sinceb0, b1, b2, . . ., bm−1 are integers, it follows thatc must be a factor of eachof the termsa0, (b0 − a1, etc. Furthermore, the very last term obtained must beequal to zero. This gives a series of checks to test whether or notc is a root. Just aswith synthetic division, any zero coefficients must be included in the algorithm. Ifcpasses each test, then not only do you know thatc is a root, but in addition you havefound the coefficients of the quotientq(x). This proves the following proposition.


1.2.4 PROPOSITION (Integer roots via backwards synthetic division).Letp(x) = amxm

+am−1xm−1+· · ·+a0 be a polynomial with integer coefficients. An

integerc is a root ofp(x) if and only if each of the following terms is an integer

b0 = −a0÷ c

b1 = (b0− a1)÷ c

b2 = (b1− a2)÷ c...

bm−1 = (bm−2− am−1)÷ c

andbm−1 = am.

Example 1.2.6.

In finding the integer roots ofp(x) = x3− 5x2

+ 2x − 10, we willuse backwards synthetic division to check the potential roots 2 and 5.(Compare what we did in Example 1.2.1.) We have

b0 = −(−10)÷ 2 = 5

b1 = (5− 2)÷ 2 = 3/2,

which eliminates 2 as a root sinceb1 is not an integer.

In testing 5 we get

b0 = −(−10)÷ 5 = 2

b1 = (2− 2)÷ 5 = 0

b2 = (0− (−5))÷ 5 = 1

andb2 = 1 = a3. At each step we got an integer value, and so 5 is aroot, giving usx3

− 5x2+ 2x − 10= (x − 5)(x2

+ 2). 2


1. Use synthetic division to writep(x) = q(x)(x − c)+ p(c) for

(a) p(x) = 2x3+ x2− 4x + 3; c = 1;

(b) p(x) = x3− 5x2

+ 6x + 5; c = 2;

(c) p(x) = x5− 7x2

+ 2; c = 2.


2. Use backwards synthetic division to show thatp(c) = 0, and to writep(x) =q(x)(x − c) for

(a) p(x) = x3− 5x2

+ 3x + 9; c = 3;

(b) p(x) = 2x5+ 2x4

− 3x2+ 2x + 5; c = −1;

(c) p(x) = x5+ 32; c = −2.

3. Use Proposition 1.2.1 to list all possible rational roots ofp(x) = 15x4+8x3

+

6x2− 9x − 2. Then determine which of the possibilities are in fact roots.

(You may eliminate possibilities graphically, using a computer or calculator ifavailable, but then verify by hand calculations that you have found the rationalroots.) Finally, use the information about roots for factorp(x) completely asa product of polynomials with integer coefficients.

4. Use the method outlined in the previous problem to find all rational roots ofp(x) = 30x5

− 7x4+ 20x3

− 4x2− 10x + 3, and use this information to

write out the factorization ofp(x).

5. Find all integer roots of the following equations (use any method). One sug-gestion is to use a calculator or computer to graph the polynomial function,then check the likely roots using synthetic division. Finally, show the com-plete factorization of each polynomial (into a product of polynomials withinteger coefficients).

(a) x3− x2− 4x − 6= 0

(b) x3− 8x2

− 3x + 91= 0

(c) x4+ 4x3

− 104x2− 105x − 108= 0

(d) x3− 6x2

− 24x + 64= 0

(e) 3x4− 5x3

− 7x2+ 9x − 2= 0

6. Let p(x) = amxm+am−1xm−1

+· · ·+a1x+a0 be a polynomial with rationalcoefficients. Show that ifb is a nonzero root ofp(x), then 1/b is a root ofq(x) = a0xm

+ a1xm−1+ · · · + am−1x + am.

1.3. APPROXIMATING REAL ROOTS 27

1.3 Approximating real roots

In this section we will develop some techniques for approximating solutions to poly-nomial equations. We will introduce the “interval bisection” method, the “secant”method, and an algorithm for approximating square roots and cube roots. In search-ing for all roots of a given equation, it is helpful to have a rough bound on the sizeof the roots. The first proposition provides such a bound.

1.3.1 PROPOSITION. Let p(x) = amxm+ · · · + a1x+ a0. If c is any root of the

equationp(x) = 0, then

|c| ≤|am| + |am−1| + · · · + |a1| + |a0|

|am|.

Proof. If c is any root of the equationp(x) = 0, then

amcm+ am−1c

m−1+ · · · + a1c+ a0 = 0,

and soamcm

= −am−1cm−1− · · · − a1c− a0.

Dividing by amcm−1 gives

c = −am−1

am

1

c− · · · −

a1

am

1

cm−2−

a0

am

1

cm−1.

If we assume that|c| ≥ 1, then1

|c|≤ 1, and so we have

|c| =

∣∣∣∣−am−1

am

1

c− · · · −

a1

am

1

cm−2−

a0

am

1

cm−1

∣∣∣∣≤

∣∣∣∣am−1

am

∣∣∣∣ ∣∣∣∣1c∣∣∣∣+ · · · + ∣∣∣∣ a1

am

∣∣∣∣ ∣∣∣∣ 1

cm−2

∣∣∣∣+ ∣∣∣∣ a0

am

∣∣∣∣ ∣∣∣∣ 1

cm−1

∣∣∣∣≤|am−1| + · · · + |a1| + |a0|

|am|.

Finally, to obtain a bound that also works when|c| < 1, we can simply add 1 to thebound we already have, giving

|c| ≤ 1+|am−1| + · · · + |a1| + |a0|

|am|

=|am| + |am−1| + · · · + |a1| + |a0|

|am|.

This completes the proof.


1.3.2 ALGORITHM (The interval bisection method for approximating roots).This algorithm is based on the following result. If we are solving the polynomialequationp(x) = 0 and find two numbersx1 < x2 for which p(x1) and p(x2) haveopposite signs, then there must be a solution betweenx1 andx2. That is, there is asolution in the interval[x1, x2].

The algorithm then proceeds as follows. To narrow down the interval in whichthere must be a solution, we evaluatep(x) at the averagex3 = (x1+ x2)/2, whichbisects the interval determined byx1 andx2. If p(x3) has the opposite sign fromp(x1), then the root is betweenx1 andx3, and so we repeat the procedure for theinterval[x1, x3]. On the other hand, ifp(x3) has the opposite sign fromp(x2), thenthe root is betweenx2 and x3, and so we repeat the procedure using the interval[x3, x2].

It is easy to write a program to carry out this algorithm. But using the intervalbisection algorithm by hand on a calculator produces numbers that are difficult towork with, and so we will use a modification of the algorithm. Suppose that theinitial interval is[0,1]. For the first step we bisect the interval, and consider either[0, .5] or [.5, 1]. Suppose thatp(.5) and p(1) have opposite signs. For the nextstep, rather than bisecting the interval, we consider either[.5, .7] or [.7, 1]. Thefirst of these can be bisected in the third step, while the interval[.7, 1] can be splitinto [.7, .8] and[.8, 1]. This gives a search procedure that increases the accuracy byone decimal point, each time that we make the necessary three or four calculations.

We will illustrate this method by solvingx2− 3 = 0, correct to three decimal

places. We know that the solution isx =√

3, which lies between 1 and 2, and sothis problem provides an easy example of the technique we want to introduce.

Let p(x) = x2− 3, so that we are solving the equationp(x) = 0. The

first three computations in the next paragraph will find thatp(1.7) = −.11 andp(1.8) = +.24, so that the answer must be between 1.7 and 1.8. It takes three morecomputations to show that the answer lies between 1.73 and 1.74, and an additionalthree computations to obtain the answer 1.732, correct to three decimal places.

After knowing that the answer is between 1 and 2, we do the computation for 1.5,half way between the two previous estimates. Becausep(1.5) = −.75 is negative,the answer must be between 1.5 and 2.0. To simplify matters and deal only withone decimal point at a time, instead of using an estimate of 1.75, which would behalf way between 1.5 and 2, we use 1.7, and then whenp(1.7) = −.11 we go top(1.8) = +.24.


The following table shows the remaining steps in the procedure.

a 1.75 1.73 1.74 1.735 1.733 1.732a2+.0625 −.0071 +.0276 +.010225 +.003289 −.000176

A graphing calculator or computer can be used in a similar way. By changing thedomain of the viewing window to give successively smaller and smaller intervals,it is possible to obtain more and more accurate estimates of where the graph of thepolynomial crosses thex-axis.

Example 1.3.1.

We will solve x4− 3 = 0, correct to five decimal places. We let

p(x) = x4− 3. (The values forp(x) have been rounded off to eight

decimal places.)

a 1.5 1.3 1.4p(a) 2.0625 −.1439 +.8416

a 1.35 1.33 1.32 1.31p(a) +.32150625 +.12900721 +.03595776 −.05500079

a 1.315 1.317 1.316p(a) −.00978090 +.00845209 −.00067480

a 1.3165 1.3163 1.3162 1.3161p(a) +.00388605 +.00206109 +.00114892 +.00023696

a 1.31605 1.31607 1.31608p(a) −.00021894 −.00003659 +.0000546

This shows that answer we are looking for is 1.31607. 2

Example 1.3.2.

In this example we will approximate a solution to the equation 2x3+

x− 2= 0. Let p(x) = 2x3+ x− 2. The possible rational solutions to

the equation arex = ±1,±2,±12, and substituting each of these values

for x shows that the solution(s) must be irrational. Butp(0) = −2 and


p(1) = 1, so there must be a solution to the equation betweenx = 0andx = 1. In the table below we have not listed all of the steps inthe search procedure. We have listed only those values for which thefunction changes sign, to show how we can narrow down the region inwhich the root lies.

x .8 .9f (x) −.176 .358

x .83 .84f (x) −.026426 .025408

x .835 .836f (x) −.0006342 .0045541

x .8351 .8352f (x) −.0001159 .0004026

x .83512 .83513f (x) −.0000121 .0000397 2

1.3.3 ALGORITHM (The secant method for approximating roots). As in theinterval bisection algorithm, to solvep(x) = 0 we begin with two numbersx1 < x2

for which p(x1) andp(x2) have opposite signs. The equation of the line joining thetwo points(x1, p(x1)) and(x2, p(x2)) is

y = m(x − x1)+ p(x1), where m=p(x2)− p(x1)

x2− x1.

If we set this equation equal to zero and solve, we get

x = x1−p(x1)

m

and then

x = x1−p(x1)(x2− x1)

p(x2)− p(x1).

Let the solution to this equation bex3, so that

x3 = x1−p(x1)(x2− x1)

p(x2)− p(x1).

We still have to findp(x3), in order to determine whether the root is in the interval[x1, x3] or in the interval[x3, x2]. We then repeat the procedure on the new interval.


To implement this algorithm on a calculator capable of storing functions withtwo variables, you can storep(x), p(y), p(z), and

f (x, y) = x − (p(x))(y− x)÷ ((p(y))− (p(x))).

Store the endpoints of the interval asx andy, and then store the new approximationasz. After evaluatingp(z) to determine its sign, you must decide whether to storeyour new approximation asx or asy, making sure that the root is in the new interval[x, y].

Example 1.3.3.

In this example we will compare the secant method with the intervalbisection method that we used in Example 1.3.2 to find a root ofp(x) =2x3+x−2 between 0 and 1. The secant method produces the following

table.

x y p(x) p(y)

0 1 −2 1.66666667 1 −.74074074 1.80851064 1 −.13445961 1.83120654 1 −.02022509 1.83455273 1 −.00295162 1.83503963 1 −.00042884 1

The table in Example 1.3.2 contains only about one third of the compu-tations actually that are actually necessary, so when we compare it withthe above table, we see that the secant method is a major improvement.

Notice that the value ofy never changed, since the successive approx-imations all remained less than the actual root. We are free to chooseany value ofy for which p(y) is positive, and so it helps to use trial anderror in finding the second endpoint. We now repeat the computations,


making appropriate modifications iny.

x y p(x) p(y)

0 1 −2 1.66666667 1 −.74074074 1.80851064 .9 −.13445961 .358.83349060 .84 −.00844657 .025408.83511467 .836 −.00003981 .00455411.83512234 .835123 −.00000003 .00000338

The table shows that to six decimal places the root is.835122. 2

Both the interval bisection method and the secant method are quite simple touse. However, for finding square roots there is another algorithm due to Newtonthat is almost as simple but much more efficient.

1.3.4 ALGORITHM (Newton’s method for approximating square roots). If ais an approximate value for

√c, then the actual value must lie betweena andc/a,

since

a ·c

a= c.

(If a = c/a, thena · a = c anda is actually equal to√

c.) If a 6= c/a, then oneof a andc/a is too large and one is too small, so the average(a+ c/a)/2 of thesetwo values will give a better approximation to

√c. This procedure, called Newton’s

method, can be continued to give more and more accurate approximations. Ifan isan approximation for

√c, then the next approximation is given by the formula

an+1 =1

2

(an +

c

an

).

If you have a calculator capable of storing functions, you can store

f (x) = (x + c÷ x)÷ 2.

If a is an approximation for√

c, then the next approximation in Newton’s methodis f (a). This makes it very easy to carry out the steps in Newton’s method.


Example 1.3.4.

We will use Newton’s method to approximate√

3 to five decimal places.We first use trial and error to get an answer accurate to one decimalplace, giving a first approximation ofa1 = 1.7. The average of 1.7 and3/1.7 is 1.7323529= a2, and then we can use this value to computethe next approximationa3. Note that the calculations have been doneon an eight digit calculator.

Step 1 Step 2 Step 3a 1.7 1.7323529 1.7320508

3/a 1.7647058 1.7317487 1.7320508avg 1.7323529 1.7320508 1.7320508

We can end with the approximationa3 = 1.7320508, since this carriesNewton’s method to the limits of the calculator.

This procedure arrives at an accurate answer much faster than the mod-ified interval bisection algorithm that we used earlier. In fact, each stepin the algorithm roughly doubles the number of decimal places to whichthe approximation is accurate. 2

Example 1.3.5.

In Newton’s method, fractions can be used instead of decimal approxi-mations. For example, squaring 7/4 gives 49/16, which is very close to3. Usinga1 = 7/4 as the first approximation to

√3 gives the sequence

of approximationsa1 = 7/4, a2 = 97/56, anda3 = 18817/10864.

To approximate√

2 we can begin witha1 = 7/5. Newton’s methodthen givesa2 = 99/70 anda3 = 19601/13860. 2


1. Use the (modified) interval bisection method to approximate3√

12 to fourdecimal places. (Work with the polynomialp(x) = x3

− 12.)

2. Use the (modified) interval bisection method to approximate√

17 to threedecimal places.


3. Use Newton’s method to approximate√

17 to six decimal places.

4. Check the calculations in Example 1.3.5.

5. Use Newton’s method to find a fraction that approximates√

17. Begin witha1 = 4, and then computea2 anda3 (as fractions).

6. Use Newton’s method to find a fraction that approximates√

10. Begin witha1 = 3, and then computea2, a3, anda4.

7. Use the (modified) interval bisection method to find the root of the equation4x3− x − 2= 0 that lies betweenx = 0 andx = 1. Show that your answer

is accurate to at least four decimal places.

8. Check the calculations in Example 1.3.3.

9. Use the secant method to solve the equation 4x3− x − 2 = 0. Obtain an

answer accurate to six decimal places.

10. Use the secant method to solve the equationx5− x2

− 6 = 0. Obtain ananswer accurate to six decimal places.

1.4. COMPLEX ROOTS 35

1.4 Complex roots

If x is any real number, thenx2+ 1 ≥ 1. This means that the equationx2

+ 1= 0has no real root. In this section we will construct a set of numbers that enlargesthe set of real numbers to include a root of this equation. A set of numbers thatcontains a rooti of the equationx2

+1= 0 and is closed under addition, subtraction,multiplication, and division must include all numbers of the forma+ bi , whereaandb are real numbers. The addition and multiplication must be given by

(a+ bi)+ (c+ di) = (a+ c)+ (b+ d)i, and

(a+ bi)(c+ di) = ac+ (bc+ ad)i + bdi2 = (ac− bd)+ (ad+ bc)i .

Here we have used the fact thati 2= −1 sincei 2

+ 1= 0.We will simply invent a symboli for which i 2

= −1, and then consider allpairs of real numbersa andb, in the forma + bi . This construction can be donemuch more formally, but at this level we hope that the reader will just accept the“invention” of the symboli .

1.4.1 DEFINITION. The setC = {a + bi | a,b ∈ R andi 2= −1} is called the

set of complex numbers, with addition and multiplication defined as follows:

(a+ bi)+ (c+ di) = (a+ c)+ (b+ d)i,

(a+ bi)(c+ di) = (ac− bd)+ (ad+ bc)i .

We say that complex numbersa+bi andc+di are equal (writtena+bi = c+di)if and only if a = c andb = d. If c+ di is nonzero, that is, ifc 6= 0 ord 6= 0, thendivision byc+ di is possible:

a+ bi

c+ di=(a+ bi)(c− di)

(c+ di)(c− di)=

ac+ bd

c2+ d2+

bc− ad

c2+ d2i .

It is often helpful to use a geometric model for the set of complex numbers.The numbera + bi can be viewed as the ordered pair(a,b) in the plane. (SeeFigure 1.4.1.) Note thati corresponds to the pair(0,1).

In polar coordinates,a+ bi is represented by(r, θ), where

r =√

a2+ b2 cosθ = a/r sinθ = b/r.


1–1

i

–i

��

a

b

a+ bisr

θ

Figure 1.4.1:

The valuer is called theabsolute valueof a+bi , and we write|a+bi | =√

a2+ b2.This definition is the same as for real numbers, since ifx ∈ R, then|x| representsthe distance fromx to 0,

The polar form for complex numbers is very useful. With this notation we have

a+ bi = r (cosθ + i sinθ).

In polar form the product of two complex numbers is given byr (cosθ + i sinθ) · t (cosφ + i sinφ)

= r t ((cosθ cosφ − sinθ sinφ)+ i (sinθ cosφ + cosθ sinφ))= r t (cos(θ + φ)+ i sin(θ + φ)).

This simplification of the product comes from the trigonometric formulas for thecosine and sine of the sum of two angles, which we recall below.

cos(θ + φ) = cosθ cosφ − sinθ sinφsin(θ + φ) = sinθ cosφ + cosθ sinφ

To multiply two complex numbers represented in polar form, we multiply theirabsolute values and add their angles. A repeated application of this formula tocosθ + i sinθ gives the following theorem.

1.4.2 THEOREM (De Moivre). For any positive integern,

(cosθ + i sinθ)n = cos(nθ)+ i sin(nθ).


1.4.3 COROLLARY. For any positive integern, the equationzn= 1 hasn distinct

roots in the set of complex numbers.

Proof. For k = 0,1, . . . ,n− 1, the values

cos2kπ

n+ i sin

2kπ

n

are distinct and(cos

2kπ

n+ i sin

2kπ

n

)n

= cos 2kπ + i sin 2kπ = 1.

Since we know thatzn= 1 has no more thann roots, we have found them all.

The complex roots ofzn= 1 are called thenth roots of unity. When plotted in

the complex plane, they form the vertices of a regular polygon withn sides inscribedin a circle of radius 1 with center at the origin.

Example 1.4.1.

The cube roots of unity are the roots of the equationx3−1= 0. We can

factor to obtain(x− 1)(x2+ x + 1) = 0, and then using the quadratic

formula we obtain the three roots

1, ω =−1

2+

√3

2i, and ω2

=−1

2−

√3

2i .

Equivalently, using Corollary 1.4.3 we obtain

1, cos2π

3+ i sin

2π

3, and cos

4π

3+ i sin

4π

3.

Note thatω2+ ω + 1 = 0, sinceω is a root of the equationz3

− 1 =(z− 1)(z2

+ z+ 1) = 0. (See Figure 1.4.2.) 2

Example 1.4.2.

The fourth roots of unity are the roots of the equationx4− 1 = 0,

which we can factor to obtain(x− 1)(x+ 1)(x2+ 1). The roots are 1,

i , i 2= −1, andi 3

= −i . (See Figure 1.4.3.) 2


1–1

i

–i

bbbb

bbb

bb

""""

""

"""

s

s

s

ω = −12 +

√3

2 i

ω2= −

12 −

√3

2 i

Figure 1.4.2: Cube roots of unity

1–1

i

–i

@@@@

@@

��

��

��

@@@@@@

ss

s

s

Figure 1.4.3: Fourth roots of unity


Example 1.4.3.

If zn= u, then(zω)n = u, whereω is anynth root of unity. Thus if all

nth roots of unity are already known, it is easy to find thenth roots ofany complex number. In general, thenth roots ofr (cosθ + i sinθ) are

r 1/n

(cos

θ + 2kπ

n+ i sin

θ + 2kπ

n

), for 1≤ k ≤ n.

To find the square root of a complex number it may be helpful to usethe formulas

sinθ

2= ±

√1− cosθ

2and cos

θ

2= ±

√1+ cosθ

2.

Example 1.4.4.

In this example we will use three different methods to find the complexcube roots of−8.

We will first use the general formula from Example 1.4.3. To do so,we need to write−8 in the polar formr (cosθ + i sinθ). Since−8 =−8+ 0 i , the corresponding rectangular coordinates are(−8,0), andsor = 8, cosθ = −1, and sinθ = 0, which gives usθ = π . Thus

3√−8= 81/3

(cos

π + 2kπ

3+ i sin

π + 2kπ

3

), for 1≤ k ≤ 3,

which gives us the three solutions

2(cosπ + i sinπ), 2

(cos

5π

3+ i sin

5π

3

), 2

(cos

π

3+ i sin

π

3

).

Further simplification gives

−2, 1−√

3 i, 1+√

3 i .

The second method, which is easier to use in this case, is to find theeasy solution3

√−8 = −2. Then the three solutions can be found by

multiplying this particular solution by the cube roots of unity

1,−1

2+

√3

2i, and

−1

2−

√3

2i .


The final method is often impractical for more complicated problems,but in this relatively easy example it uses more familiar techniques. Tofind the cube roots of−8 we need to solve the equationx3

+ 8 = 0.We can factor, to obtainx3

+ 8= (x + 2)(x2− 2x + 4). Then we can

use the quadratic formula to find the roots ofx2−2x+4, giving us the

same solutions

−2, 1−√

3 i, 1+√

3 i .

We have noted that the powers ofi are i , i 2= −1, i 3

= −i , i 4= 1. Since

i 4= 1, the powers repeat. For example,i 5

= i 4i = i , i 6= i 4i 2

= −1, and so on.For any integern, the poweri n depends on the remainder ofn when divided by 4,since ifn = 4q + r , theni n

= i 4q+r= (i 4)qi r

= i r . In particular,i−1= i 3= −i ,

and a similar computation can be given for any negative exponent.The next theorem is usually referred to as the “fundamental theorem of algebra.”

It was discovered by D’Alembert in 1746, although he gave an incorrect proof. Thefirst acceptable proof was given by Gauss in 1799. The proof is beyond the scopeof this text.

1.4.4 THEOREM (The fundamental theorem of algebra).Every polynomial ofpositive degree with complex coefficients has a complex root.

1.4.5 COROLLARY. Every polynomialf (z) of degreen > 0 with complex coef-ficients can be expressed as a product of linear factors, in the form

f (z) = c(z− z1)(z− z2) · · · (z− zn).

Proof. To give a proof of the corollary, we only need to combine the FundamentalTheorem of Algebra with the fact that roots off (z) correspond to linear factors.

If z= a+ bi is a complex number, then itscomplex conjugate, denoted byz,is z = a− bi . Note thatzz = a2

+ b2 andz+ z = 2a are real numbers, whereasz− z= (2b)i is a purely imaginary number. Furthermore,z= z if and only if z is areal number ( i.e.,b = 0). It can be checked that(z+ w) = z+w and(zw) = zw.

1.4.6 PROPOSITION. Let f (x) be a polynomial with real coefficients. Then acomplex numberz is a root of f (x) if and only if z is a root of f (x).


Proof. If f (x) = anxn+ · · · + a0, thenanzn

+ · · · + a0 = 0 for any rootz of f (x).Taking the complex conjugate of both sides shows that

an(z)n+ · · · + a1z+ a0 = an(z)

n+ · · · + a1z+ a0 = 0

and thusz is a root of f (x). Conversely, ifz is a root of f (x), then so isz= z.

1.4.7 THEOREM. Any polynomial with real coefficients can be factored into aproduct of linear and quadratic terms with real coefficients.

Proof. Let f (x) be a polynomial with real coefficients, of degreen. By Corol-lary 1.4.5 we can writef (x) = c(x − z1)(x − z2) · · · (x − zn), wherec ∈ R. If zi

is not a real root, then by Proposition 1.4.6,zi is also a root, and sox− zi occurs asone of the factors. But then

(x − zi )(x − zi ) = x2− (zi + zi )x + zi zi

has real coefficients. Thus if we pair each nonreal root with its conjugate, theremaining roots will be real, and sof (x) can be written as a product of linear andquadratic polynomials each having real coefficients.

1.4.8 COROLLARY. Any polynomial of odd degree that has real coefficients musthave a real root.

Proof. By the previous theorem, such a polynomial must have a linear factor withreal coefficients, and this factor yields a real root.


1. Compute each of the following:

(a) (−1/√

2+ i /√

2)4

(b) (−1+ i )8

(c) (cos 30◦ + i sin 30◦)15

2. Find(a+ bi)−1, if a+ bi lies on the unit circle.

3. (a) Find the 6th roots of unity.

(b) Find the 8th roots of unity.


4. (a) Find the cube roots of−8i .

(b) Find the fourth roots of−1.

5. Solve the equationx3− x2− 4= 0.

6. Solve the equationx3+ 3x2

− 6x + 20= 0.

7. Prove that the sum of the roots of the equationax2+ bx+ c = 0 is−b/a,

and that the product of the roots isc/a.

8. Use DeMoivre’s theorem to find formulas for cos 3θ (in terms of cosθ ) andsin 3θ (in terms of sinθ ).

9. Verify each of the following, for complex numbersz andw:

(a) zz= |z|2

(b) zw = zw

(c) |zw| = |z||w|

(d) z= z if and only if z ∈ R

10. Leta andb be integers, each of which can be written as the sum of two perfectsquares. Show thatab has the same property.

Hint: Use part (c) of the previous problem.

1.5. INTERPOLATING POLYNOMIALS 43

1.5 Interpolating polynomials

Most “real world” problems involve a function of some type. Calculus textbooksusually assume that we either already have a formula or can easily find one. Butin practice we often have only a few data points, and have to find a formula thatfits the data. If we are able to predict the type of formula to use, then the problemis to find the formula of the given type that gives the closest fit to the data. In thissection we will deal with a simpler problem. We assume that the formula we needis a polynomial, and that it fits the data exactly.

Two points in a plane determine a unique line that passes through them. Wewill show that given three points, which do not all lie on the same line, it is possibleto find a parabola that passes through them. It is easiest for us to work with graphsof functions, and so we must assume that no two points lie on the same verticalline. This means that we can assume that the parabola determined by three pointsis actually represented by the graph of a quadratic polynomial.

For a larger number of points, what we will show is that given distinct numbersx0, x1, . . . , xn together with numbersy0, y1, . . . , yn (which need not be distinct),there is a polynomialp(x) of degree at mostn such thatp(xi ) = yi for each indexi .There is at most one such polynomial of degreen (as we showed in Proposition 1.1.8).

We begin with a familiar case. Given points(x0, y0) and(x1, y1), the two-pointform of the equation of the line through the given points is

y− y0

x − x0=

y1− y0

x1− x0.

This can be simplified to give

y = y0+y1− y0

x1− x0(x − x0).

A formula that begins with these terms will provide a basis for one solution toour general problem. It is difficult to write down in general, and so we first giveanother solution whose formula is easier to remember. We can rewrite the equationas follows:

y = y0+ y1(x − x0)

(x1− x0)− y0

(x − x0)

(x1− x0)

= y0(x0− x1)

(x0− x1)+ y0

(x − x0)

(x0− x1)+ y1

(x − x0)

(x1− x0)

= y0(x − x1)

(x0− x1)+ y1

(x − x0)

(x1− x0)


It is important to understand how the above function is constructed. As a functionof x it has the form

f (x) = f0(x)+ f1(x),

where f0(x0) = y0, f0(x1) = 0, and f1(x0) = 0, f1(x1) = y1.To solve the problem of finding a polynomial function that passes through three

given points(x0, y0), (x1, y1), (x2, y2), we will define three polynomialsf0(x),f1(x), and f2(x) such that

f0(x0) = y0 f1(x0) = 0 f2(x0) = 0f0(x1) = 0 f1(x1) = y1 f2(x1) = 0f0(x2) = 0 f1(x2) = 0 f2(x2) = y2

Then f (x) = f0(x)+ f1(x)+ f2(x) has the properties we want:f (x0) = y0+0+0,f (x1) = 0+ y1+ 0, and f (x2) = 0+ 0+ y2.

How do we constructf0(x), f1(x), and f2(x)? Sincef0(x1) = 0 and f0(x2) = 0,we can see thatf0(x) must havex − x1 andx − x2 as factors, so we assume thatf0(x) has the form

f0(x) = k(x − x1)(x − x2)

wherek is a constant. Sincef0(x0) = y0, we can substitute and solve fork to get

k =y0

(x0− x1)(x0− x2).

We can findf1(x) and f2(x) similarly. This gives a derivation of the formula in thefollowing proposition.

1.5.1 PROPOSITION. Let x0, x1, x2 be distinct real numbers, and lety0, y1, y2

be any real numbers. Then the polynomial

f (x) =y0(x − x1)(x − x2)

(x0− x1)(x0− x2)+

y1(x − x0)(x − x2)

(x1− x0)(x1− x2)+

y2(x − x0)(x − x1)

(x2− x0)(x2− x1)

has the property thatf (x0) = y0, f (x1) = y1, and f (x2) = y2.

Example 1.5.1.

To find the polynomial of degree 2 whose graph passes through thepoints(−1,1), (0,0), and(1,1), we can use the Lagrange interpolationformula. We have

f (x) =1(x − 0)(x − 1)

(−1− 0)(−1− 1)+

0 (x + 1)(x − 1)

(0+ 1)(0− 1)+

1(x + 1)(x − 0)

(1+ 1)(1− 0),


and simplifying givesf (x) = 12x(x− 1)+ 1

2(x+ 1)x. Usually we areinterested in substituting values ofx close to one of the given points,and so it makes sense to leave the formula as is, rather than rewriting itin terms of powers ofx, as f (x) = x2. 2

Givenn+ 1 points, the Lagrange interpolation formula can easily be extendedto find the polynomial of degreen (or less) whose graph passes through the points.In order to be able to state the general formula in a reasonable way we need tointroduce some notation. Just as we use

∑to denote sums, we will use

∏to denote

products. The next theorem uses this notation to state (without proof) the generalLagrange interpolation formula.

1.5.2 THEOREM (The Lagrange interpolation formula). Let x0, x1, . . ., xn

be distinct real numbers, and lety0, y1, . . ., yn be any real numbers. Then thepolynomial

p(x) =n∑

i=0

yi∏

s6=i (x − x j )∏s6=i (xi − x j )

has the property thatf (xi ) = yi , for i = 0, . . . ,n.

One difficulty with the Lagrange interpolation formula is that if an additionalpoint is given, then everything must be recomputed. To motivate another approach,using what are termeddivided differences, we go back to the two point form of astraight line, as given below.

f (x) = y0+y1− y0

x1− x0(x − x0).

In this case our function is expressed as a sum of two functionsf0(x) and f1(x)such thatf0(x0) = y0 and f1(x0) = 0, while f0(x1) = y0 and f1(x1) = y1 − y0.(Compare this with Lagrange’s formula, in whichf0(x0) = y0 and f1(x0) = 0,while f0(x1) = 0 and f1(x1) = y1.)

To add a third termf3(x) so thatf (x)would define a quadratic function passingthrough(x0, y0), (x1, y1), and(x2, y2), we look for three polynomialsf0(x), f1(x),and f2(x) such that

f (x) = f0(x)+ f1(x)+ f2(x),

where f0(x) has degree zero,f1(x) has degree one, andf2(x) has degree two. Thepolynomials f0(x), f1(x), and f2(x) will be chosen so that


f0(x0) = y0 f1(x0) = 0 f2(x0) = 0f0(x1) = y0 f1(x1) = y1− f0(x1) f2(x1) = 0f0(x2) = y0 f1(x2) = f1(x2) f2(x2) = y2− f1(x2)− f0(x2)

Since f2(x0) = 0 and f2(x1) = 0, we look for a quadratic function of the form

f2(x) = k(x − x0)(x − x1).

Since we must havef (x2) = y2, we can determine the constantk by just substitutingx = x2 into the above equation. We obtain

k =

y2− y1

x2− x1−

y1− y0

x1− x0

(x2− x0).

This leads to the equation

f (x) = y0+y1− y0

x1− x0(x − x0)+ k(x − x0)(x − x1).

Notice that the first term in the numerator ofk is the slope of the line segment joining(x1, y1) and(x2, y2), while the second term is the slope of the line segment joining(x0, y0) and(x1, y1), which we have already computed. This is called thedivideddifferencesinterpolation formula.

Example 1.5.2.

We will use the divided differences method to determine a quadraticthrough the points(1,1), (2,4), and(3,9), knowing that we shouldobtain the functionf (x) = x2. We letx0 = 1, x1 = 2, andx2 = 3.Then the function we are looking for has the form

f (x) = a+ b(x − 1)+ c(x − 1)(x − 2),

where

a = 1,b =4− 1

2− 1= 3, andc =

9− 4

3− 2−

4− 1

2− 13− 1

=5− 3

3− 1= 1.

This gives us the polynomialf (x) = 1+ 3(x− 1)+ 1(x− 1)(x− 2).It can be used in this form, without combining terms. (You can check


that it reduces tof (x) = x2, if you like.) For example,f (2.5) =1+ 3(1.5)+ (1.5)(.5) = 6.25.

Note that the computation ofc uses the value obtained forb, togetherwith another value computed similarly. To simplify the computations,it is convenient to arrange the necessary terms in a table in the followingway. Here each column of divided differences is constructed from theprevious one. Then the coefficients of the polynomial are found byreading from left to right, along the bottom of each column.

x y3 9

9−43−2 = 5

2 4 5−33−1 = 1

4−12−1 = 3

1 1

Example 1.5.3.

In this example we will find a polynomial of degree at most 5 whosegraph passes through the points

(3,19), (2.5,16.125), (1,3), (.5,−.0375), (0,2), and(−1,3).

We begin the table of divided differences by listing thex-values indescending order, together with the correspondingy-values. Listingthex-values in descending order tends to simplify the computations.

x y3.0 19.000

5.752.5 16.125 −1.5

8.75 −11.0 3.000 1.0 0

6.75 −1 00.5 −0.375 3.5 0

3.25 −10.0 −2.000 5.5

−5.00−1.0 3.000

For example, the value 5.5 in the fourth column is computed by sub-tracting−5.00 from 3.25, the immediately preceding values in the table.


Since this is in the second column of differences, the denominator isthe difference of thex values 0.5 and−1.0.

In the next column, the bottom term−1 is computed by subtracting 5.5from 3.5 and then dividing by the difference of thex values 1.0 and−1.0.

The zeros in the table show that only four coefficients are necessary,and so the six given points actually lie on a cubic. Reading off thecoefficients from the bottoms of the columns, we have 3,−5, 5.5,−1.The first of these is the constant term. The next coefficient correspondsto the lastx value in the table, which is−1, and so the term isx−(−1) =x + 1. The next coefficient corresponds to the term given by the lasttwo x values, namely(x − 1)(x − 0). Continuing in this way gives usthe polynomial

f (x) = 3− 5(x + 1)+ 5.5(x + 1)(x)− (x + 1)(x)(x − .5).

Example 1.5.4.

We know that 1+ 2+ · · · + n = n(n + 1)/2. This suggests that theformula for the sum of squares might be a cubic polynomial inn. Letus write p(n) =

∑ni=0 i 2. Then p(0) = 0, p(1) = 02

+ 12= 1,

p(2) = 02+ 12+ 22

= 5, andp(3) = 02+ 12+ 22+ 32

= 14. Tomake an educated guess at the general formula, we can use the Lagrangeinterpolation formula to get

p(n) =0 (n− 1)(n− 2)(n− 3)

(0− 1)(0− 2)(0− 3)+

1(n− 0)(n− 2)(n− 3)

(1− 0)(1− 2)(1− 3)

+5(n− 0)(n− 1)(n− 3)

(2− 0)(2− 1)(2− 3)+

14(n− 0)(n− 1)(n− 2)

(3− 0)(3− 1)(3− 2).

After simplifying (this is left as an exercise) we get

p(n) = n(n+ 1)(2n+ 1)/6 .

If we use the divided differences technique, we have the following tableof differences.


n p(n)3 14

92 5 5/2

4 1/31 1 3/2

10 0

This gives the polynomialp(n) = 0+ 1(n− 0)+ 32(n− 0)(n− 1)+

13(n− 0)(n− 1)(n− 2), which again can be simplified to

p(n) = n(n+ 1)(2n+ 1)/6 .


1. Use the Lagrange interpolation formula to find the polynomial of degree 2whose graph passes through(0,5), (1,7), and(−1,9).

2. Extend the Lagrange interpolation formula to the case of 4 points. Use yourformula to find the cubic polynomial whose graph passes through the points(0,−5), (1,−3), (−1,−11), and(2,1).

3. Do the first problem using divided differences.

4. Do the second problem using divided differences.

5. The following values give√

x to 6 decimal place accuracy:

(55, 7.416198), (54, 7.348469), (53, 7.280110),

(52, 7.211103), (51, 7.141428), (50, 7.071068).

(a) Compute the entire table of divided differences.

(b) Find the interpolating polynomial of degree 5 for the given values (do notsimplify) and use it to approximate

√50.25.

6. Fill in the missing algebraic steps in simplifying the two polynomials inExample 1.5.4.

7. Use divided differences to guess a formula for the sum ofn cubes.


1.6 Mathematical Induction

The principle of mathematical induction applies to statements that involve an arbi-trary positive integern. Examples of such statements are:

1. x − 1 is a factor ofxn− 1;

2. (cosθ + i sinθ)n = cos(nθ)+ i sin(nθ);

3. 1+ 2+ · · · + n = n(n+ 1)/2;

4. 12+ 22+ · · · + n2

= n(n+ 1)(2n+ 1)/6;

5. 13+ 23+ · · · + n3

= n2(n+ 1)2/4;

6. 3 is a factor of 10n − 1;

7. n2− n+ 41 is a prime number.

Notice that each statement depends on the positive integern and becomes eithertrue or false when some value is substituted forn. The above statements are true,except for the last one:n2

− n+ 41 is a prime number whenn = 1,2, . . . ,10, butnot whenn = 41.

Suppose thatPn is a statement depending on the positive integern. If Pn istrue for each choice ofn, then the principle of mathematical induction frequentlyallows us to establish this fact. We will state the principle, and then apply it to someexamples. Note that we could begin numbering with any integer, sayP0, P1, . . . orevenP−297, P−296, . . ..

1.6.1 THEOREM (Principle of Mathematical Induction). Let P1, P2, . . . be asequence of propositions. Suppose that

(i) P1 is true, and

(ii) if Pk is true, thenPk+1 is true for all positive integersk.ThenPn is true for all positive integersn.

We will illustrate the use of induction by giving some examples and provingtwo theorems.

1.6. MATHEMATICAL INDUCTION 51

Example 1.6.1.

As our first example, we will prove by induction thatx − 1 is a factorxn− 1, for all positive integersn. We should note that this result is a

special case of Lemma 1.1.3, but at that point in the text we gave aninformal proof that avoided using mathematical induction.

Let Pn be the statement thatx − 1 is a factor ofxn− 1. It is obvious

that theP1 is true. Next we will assume thatPk is true and prove thatPk+1 is true. We can write

xk+1− 1= xk+1

− xk+ xk− 1 .

Thenxk+1− xk

= (x − 1)xk, and since we have assumed thatx − 1is a factor ofxk

− 1, we have writtenxk+1− 1 as a sum of two terms,

each of which hasx − 1 as a factor. This shows thatx − 1 is a factorof xk+1

− 1, and thereforePk+1 is true. This completes the inductionproof. 2

This is a good point at which to emphasize that when we are using the principleof mathematical induction, we must establish the truth ofP1. However, when weestablish the truth ofPk+1, we get to assume the truth ofPk without having to proveanything about it.

Example 1.6.2.

To establish that 1+ 2+ · · · + n = n(n+ 1)/2, let Pn be the statement1+ 2+ · · · + n = n(n+ 1)/2. Whenn = 1 we have 1= 1(1+ 1)/2,so P1 is true.

The next step is to show thatPk implies Pk+1. Assume thatPk is true,so that we have 1+ 2+ · · · + k = k(k + 1)/2. Addingk + 1 to bothsides of this equation, we get

1+ 2+ · · · + k+ (k+ 1) =k(k+ 1)

2+ (k+ 1)

=k(k+ 1)+ 2(k+ 1)

2

=(k+ 1)(k+ 2)

2

=(k+ 1)[(k+ 1)+ 1]

2.


ThusPk+1 is true, and so by inductionPn holds for all positive integersn. 2

Example 1.6.3.

To establish that 13 + 23+ · · · + n3

= n2(n + 1)2/4, let Pn be thestatement 13+ 23

+ · · · + n3= n2(n+ 1)2/4. Then 1= 12(1+ 1)2/4,

so P1 is true. The next step is to show thatPk implies Pk+1. AssumethatPk is true, so that we have 13

+23+· · ·+ k3

= k2(k+1)2/4. Add(k+ 1)3 to both sides of this equation to get

13+ 23+ · · · + k3

+ (k+ 1)3 =k2(k+ 1)2

4+ (k+ 1)3 .

We must show that the right hand side of the equation gives us theformula in Pk+1, which is

(k+ 1)2(k+ 2)2

4.

That part of the proof is left to the reader, as Problem 3. 2

Example 1.6.4.

To prove that 3 is a factor of 10n− 1 for all positive integersn, let Pn

be the statement that 3 is a factor of 10n−1. Now P1 is true since 3 is a

factor of 9. Assume thatPk is true, that is, that 3 is a factor of 10k− 1.

Then since

10k+1−1= 10·10k

−1= 10·10k−10+10−1= 10· (10k

−1)+9

we see that 3 is a factor of 10k+1− 1 since it is a factor of both 10k

− 1and 9. HencePk+1 is true. By the principle of mathematical induction,Pn holds for all positive integersn. 2

We next give a proof by induction of Theorem 1.4.2.

1.6.2 THEOREM (De Moivre). For any positive integern,

(cosθ + i sinθ)n = cos(nθ)+ i sin(nθ) .


Proof. To give a proof by induction, we first observe that there is nothing to provewhenn = 1. Now assume that the statement is true forn = k. Then

(cosθ + i sinθ)k+1= (cosθ + i sinθ)k(cosθ + i sinθ)

= (cos(kθ)+ i sin(kθ))(cosθ + i sinθ)

= (cos(kθ) cosθ − sin(kθ) sinθ)

+i (sin(kθ) cosθ + cos(kθ) sinθ))

= cos(kθ + θ)+ i sin(kθ + θ))

= cos((k+ 1)θ)+ i sin((k+ 1)θ) .

The crucial step uses two trig identities: cos(φ+ θ) = cosφ cosθ − sinφ sinθ andsin(φ + θ) = sinφ cosθ + cosφ sinθ .

Before proving the Binomial Theorem, we need to introduce some notation and

prove a Lemma. The symbol

(n

i

)is called abinomial coefficient, and it is referred

to as “n choosei ”. It gives the number of ways in whichi elements can be chosenout of a set withn elements.

There aren ways to choose the first element,n− 1 ways to choose the secondelement out of the remainingn− 1 elements, and so on until the last step, at whichpoint there aren − (i − 1) ways to choose theith element out of the remainingn−(i −1) elements. To find the correct number of possibilities that are independentof the order in which the elements are chosen, we must divide by the number ofways in which thei elements can be reordered. This gives us(

n

i

)=

n(n− 1) · · · (n− i + 1)

i (i − 1) · · · 1

=n(n− 1) · · · (n− i + 1)(n− i ) · · · 1

i (i − 1) · · · 1(n− i ) · · · 1

=n!

i !(n− i )!,

wheren! = n(n− 1) · · · 2 · 1 for n ≥ 1 and 0! = 1.

1.6.3 LEMMA. For any positive integerk and any positive integeri with i ≤ k,(k

i

)+

(k

i − 1

)=

(k+ 1

i

).


Proof. Writing out the definition of the binomial coefficients and finding a commondenominator for the sum, we have(

k

i

)+

(k

i − 1

)=

k!

i !(k− i )!+

k!

(i − 1)!(k− i + 1)!

=(k− i + 1)k!

i !(k− i )!(k− i + 1)+

(i )k!

(i )(i − 1)!(k− i + 1)!

=(k− i + 1)k!

i !(k− i + 1)!+

(i )k!

(i )!(k− i + 1)!

=(k+ 1)k!

i !(k+ 1− i )!

=

(k+ 1

i

).

This completes the proof.

1.6.4 THEOREM (The Binomial Theorem). Let a andb be real numbers. Thenfor any positive integern,

(a+ b)n =n∑

i=0

(n

i

)an−i bi .

Proof. Let Pn be the statement that(a+ b)n =n∑

i=0

(n

i

)an−i bi . Then

1∑i=0

(1

i

)a1−i bi

=

(1

0

)a1b0+

(1

1

)a0b1= a+ b

and soP1 is true. Next, assume thatPk is true. Then we have

(a+ b)k+1= (a+ b)(a+ b)k

= (a+ b)

(k∑

i=0

(k

i

)ak−i bi

)

=

k∑i=0

(k

i

)ak−i+1bi

+

k∑i=0

(k

i

)ak−i bi+1 .


Next we break the two sums apart to get

(a+ b)k+1=

(k

0

)ak+1b0

+

k∑i=1

(k

i

)ak+1−i bi

+

k−1∑i=0

(k

i

)ak−i bi+1

+

(k

k

)a0bk+1 .

Then we can change the indices in the summations, so that we can combine termsby using Lemma 1.6.3.

(a+ b)k+1=

(k

0

)ak+1b0

+

k∑i=1

(k

i

)ak+1−i bi

+

k∑i=1

(k

i − 1

)ak−i+1bi

+

(k

k

)a0bk+1

= ak+1b0+

k∑i=1

[(k

i

)+

(k

i − 1

)]ak+1−i bi

+ a0bk+1

=

(k+ 1

0

)ak+1b0

+

k∑i=1

(k+ 1

i

)ak+1−i bi

+

(k+ 1

k+ 1

)a0bk+1

=

k+1∑i=0

(k+ 1

i

)ak+1−i bi .

ThusPk+1 is true, and the proof is complete.

A second form of mathematical induction is more useful for some purposes.

1.6.5 THEOREM. (Second Principle of Mathematical Induction) LetP1, P2, . . .

be a sequence of propositions. Suppose that(i) P1 is true, and(ii) if Pm is true for allm≤ k, thenPk+1 is true for allk ∈ Z+.

ThenPn is true for all positive integersn.

Example 1.6.5.

Define a sequence of natural numbers as follows: LetF1 = F2 = 1, andFn = Fn−1 + Fn−2 for n ≥ 3. ThusF3 = 2, F4 = 3, F5 = 5, F6 = 8,


etc. The sequenceF1, F2, . . . is called theFibonacci sequence. Wewill show thatFn < (7/4)n for all positive integersn.

Let Pn be the statementFn < (7/4)n. ThenP1 saysF1 = 1< (7/4)1,which is true. Assuming thatPm for all m≤ k, we have

Fk+1 = Fk + Fk−1

<

(7

4

)k

+

(7

4

)k−1

=

(7

4

)k−1(7

4+ 1

)=

(7

4

)k−1(11

4

)<

(7

4

)k−1(49

16

)=

(7

4

)k+1

,

and soPk+1 holds. Therefore, by the second principle of mathematicalinduction, we haveFn < (7/4)n for all positive integersn. 2


Use the principle of mathematical induction to establish each of the following, wheren is any positive integer:

1. n < 2n

2. 12+ 22+ · · · + n2

= n(n+ 1)(2n+ 1)/6

3. 13+ 23+ · · · + n3

= n2(n+ 1)2/4 (Complete the proof in Example 1.6.3).

4. 1+ 3+ 5+ · · · + (2n− 1) = n2

5. 2+ 22+ · · · + 2n

= 2n+1− 2

6. x − 1 is a factor ofxn− 1

7. x + 1 is a factor ofx2n−1+ 1


8. 1+ r + r 2+ · · · + r n

=1− r n

1− r(whenr 6= 1)

9.n∑

i=1

1

i · (i + 1)=

n

n+ 1

10.n∑

i=0

(n

i

)= 2n

Hint: Use the fact that(k+1

i

)=(k

i

)+( k

i−1

).

the calculus of polynomialsbeachy/courses/229/03fall/calculus.pdf · preface these notes are...

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